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A line L is a drawn from P(4,3) to meet ...

A line `L` is a drawn from `P(4,3)` to meet the lines `L-1a n dL_2` given by `3x+4y+5=0` and `3x+4y+15=0` at points `Aa n dB` , respectively. From `A` , a line perpendicular to `L` is drawn meeting the line `L_2` at `A_1dot` Similarly, from point `B_1dot` Thus, a parallelogram `AA_1B B_1` is formed. Then the equation of `L` so that the area of the parallelogram `AA_1B B_1` is the least is `x-7y+17=0` `7x+y+31=0` `x-7y-17=0` `x+7y-31=0`

Text Solution

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The given lines `(L_(1) " and" L_(2))` are parallel and the distance between them (BC or AD) is (15-5)/5 = 0 units. Let `angle BAC = theta.` So, AB = BC cosec `theta = 2` cosec `theta " and AA"_(1) = AD " sec " theta = 2 " sec " theta.` Now, the area of parallelogram `"AA"_(1)BB_(1)` is
`Delta = AB xx "AA"_(1) = 4 "sec " theta " cosec" theta`
` = (8)/("sin" 2 theta)`
` " Clearly", Delta " is the least for" theta = pi//4. " Let the slope of AB be m. Then,"`
`tan45^(@) = |(m+3//4)/(1-(3m//4))|`
` " or " 4m+3= +-(4-3m) or m = (1)/(7) or -7`
Hence, the equation of L is
x-7y+17=0
or 7x + y -31=0
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